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# July 2003 Puzzle Solution [Preview]

In the first problem, one container must travel from B to A and one from A to B, because no containers at all are coming to A or B from C, D or E. Therefore, three containers from A and one container from B are going to C, D or E. None of these four containers can travel to C, because then the two coming from D and E would have nowhere to go. So the six containers going to D and E consist of three from A, one from B and two from C. Thus, D gets at least one and at most three from A. E gets at most two and possibly none.

In the second problem, it is possible for A to send two containers to B. In that case, two containers from A, two from B and two from C head to D or E. So 3 or fewer can go from A to D and 2 or fewer from A to E. (Thanks to reader Louis Tagliaferro for pointing out an error in my earlier answer.)

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